#### Answer

$x = 2~cos~2t$
$y = 2~sin~2t$
$t$ in $[0,2\pi]$
We can see the graph below:

#### Work Step by Step

$x = 2~cos~2t$
$y = 2~sin~2t$
$t$ in $[0,2\pi]$
When $t = 0$:
$x = 2~cos~0 = 2$
$y = 2~sin~0 = 0$
When $t = \frac{\pi}{8}$:
$x = 2~cos~\frac{\pi}{4} = 2~\frac{\sqrt{2}}{2} = \sqrt{2}$
$y = 2~sin~\frac{\pi}{4} = 2~\frac{\sqrt{2}}{2} = \sqrt{2}$
When $t = \frac{\pi}{4}$:
$x = 2~cos~\frac{\pi}{2} = 0$
$y = 2~sin~\frac{\pi}{2} = 2$
When $t = \frac{\pi}{3}$:
$x = 2~cos~\frac{2\pi}{3} = -1$
$y = 2~sin~\frac{2\pi}{3} = \sqrt{3}$
When $t = \frac{\pi}{2}$:
$x = 2~cos~\pi = -2$
$y = 2~sin~\pi = 0$
When $t = \frac{3\pi}{4}$:
$x = 2~cos~\frac{3\pi}{2} = 0$
$y = 2~sin~\frac{3\pi}{2} = -2$
When $t = \pi$:
$x = 2~cos~2\pi = 2$
$y = 2~sin~2\pi = 0$
As $t$ increases from $t = \pi$ to $t=2\pi$, the points around the circle of radius 2 are repeated.
We can see the graph below: